Answer:
db / dt = kb
this becomes b(t) = Ce^(kt)
C = 100, the initial population
P(1) = 420 = 100 e^(1k)
4.2 = e^k
ln 4.2 = k
a) thus, b(t) = 100 e^(t ln 4.2)
b) b(3) = 100 e^(3 ln 4.2)
c) growth constant will still be ln 4.2 (constant percentage of populatioin)
d) 10000 = 100 e^(t ln 4.2)
100 = e^(t ln 4.2)
ln 100 = t ln 4.2
t = ln 100 / ln 4.2
Step-by-step explanation:
Answer:
The probability that the sample mean will lie within 2 values of μ is 0.9544.
Step-by-step explanation:
Here
- the sample size is given as 100
- the standard deviation is 10
The probability that the sample mean lies with 2 of the value of μ is given as

Here converting the values in z form gives

Substituting values

From z table

So the probability that the sample mean will lie within 2 values of μ is 0.9544.
Answer:
y = 0
Step-by-step explanation: